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27 Oct 2016, 23:37
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Guys I was wondering how to solve ( basically it is a factorisation issue) this.
10x^2-19x+6 ≤0
Answer is 2/5≤x≤3/2 but my issue is factoring the quadratic.
10x^2-19x+6 ≤0
Answer is 2/5≤x≤3/2 but my issue is factoring the quadratic.
27 Oct 2016, 23:57
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Bookmarks
I'll give it a try and see if it helps you.
ax^2 + bx + c is the current form of the equation.
In order to solve this we have to find factors which one multiplication give you a*c and on addition give you b (-ve or +ve sign of b will result in which of factor is bigger)
So in the question above we have
10x^2-19x+6 = 0 (For simplification we shall work with = 0 as of now)
so a = 10 and c = 6 so a*c = 60
b = -19
Factors of a*c = 1,2,3,4,5,6,10,12,15,20,30,60
from these factors we have to find two of the which on multiplication will give 60 and on addition/subtraction will give you 19. We know that if we multiply first factor and last factor we get 60. Similarly if we multiply 2nd factor and 2nd last factor we get 60.(you can refer to GMAT club math's book for this concept)
so taking two factors at a time 2*30 = 60 but no matter if we add them or subtract them we won't get 19 we can only get 28 or 32. So, this pair doesn't help us.
You can try with others and realize that 15 and 4 work for this equation, as 15*4 = 60 and adding 15 and 4 will give you 19.
Now if we see b is having a negative sign and a*c is positive so both the factors (15 and 4) will have to have negative sign. As both negatives will add up and maintain the -ve sign and on multiplication of -ve*-ve will result in +ve sign.
So the above equation can be written as
10x^2 -15x -4x + 60 = 0
5x(2x-3) -2(2x-3) = 0
(2x -3)(5x - 2) = 0
HTH.
Let me know if you have any questions on this.
ax^2 + bx + c is the current form of the equation.
In order to solve this we have to find factors which one multiplication give you a*c and on addition give you b (-ve or +ve sign of b will result in which of factor is bigger)
So in the question above we have
10x^2-19x+6 = 0 (For simplification we shall work with = 0 as of now)
so a = 10 and c = 6 so a*c = 60
b = -19
Factors of a*c = 1,2,3,4,5,6,10,12,15,20,30,60
from these factors we have to find two of the which on multiplication will give 60 and on addition/subtraction will give you 19. We know that if we multiply first factor and last factor we get 60. Similarly if we multiply 2nd factor and 2nd last factor we get 60.(you can refer to GMAT club math's book for this concept)
so taking two factors at a time 2*30 = 60 but no matter if we add them or subtract them we won't get 19 we can only get 28 or 32. So, this pair doesn't help us.
You can try with others and realize that 15 and 4 work for this equation, as 15*4 = 60 and adding 15 and 4 will give you 19.
Now if we see b is having a negative sign and a*c is positive so both the factors (15 and 4) will have to have negative sign. As both negatives will add up and maintain the -ve sign and on multiplication of -ve*-ve will result in +ve sign.
So the above equation can be written as
10x^2 -15x -4x + 60 = 0
5x(2x-3) -2(2x-3) = 0
(2x -3)(5x - 2) = 0
HTH.
Let me know if you have any questions on this.
